Let the circle $x^2+y^2=4$ intersect $x$-axis at the points $\mathrm{A}(\mathrm{a}, 0), \mathrm{a}>0$ and $\mathrm{B}(\mathrm{b}, 0)$. Let $\mathrm{P}(2 \cos \alpha, 2 \sin \alpha), 0<\alpha<\frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha-\beta)=\frac{\pi}{2}$. Then the point of intersection of AQ and BP lies on:
